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3 years ago

Ok we trying this again. so just in case if the picture isnt showing, here are the questions:

Chemistry

1 answer:

11Alexandr11 [23.1K]3 years ago

3 0

Answer:

1. 31.68 moles of water, H₂O

2. 14.81 moles of Cr

Explanation:

1. Determination of the number of mole of water, H₂O.

The balanced equation for the reaction is given below:

C₁₂H₂₂O₁₁ —> 12C + 11H₂O

From the balanced equation above,

1 mole of C₁₂H₂₂O₁₁ produced 12 moles of C and 11 moles of H₂O.

Next, we shall determine the number of mole C₁₂H₂₂O₁₁ needed to produce 34.55 moles of C. This can be obtained as follow:

From the balanced equation above,

1 mole of C₁₂H₂₂O₁₁ produced 12 moles of C.

Therefore, Xmol of C₁₂H₂₂O₁₁ will produce 34.55 moles of C i.e

Xmol of C₁₂H₂₂O₁₁ = 34.55 / 12

Xmol of C₁₂H₂₂O₁₁ = 2.88 moles

Thus, 2.88 moles of C₁₂H₂₂O₁₁ is needed.

Finally, we shall determine the number of mole of water, H₂O produced from the reaction. This can be obtained as follow:

From the balanced equation above,

1 mole of C₁₂H₂₂O₁₁ produced 11 moles of H₂O.

Therefore, 2.88 moles of C₁₂H₂₂O₁₁ will produce = 2.88 × 11 = 31.68 moles of H₂O.

Thus, 31.68 moles of water, H₂O were obtained from the reaction.

2. Determination of the number of mole of Cr needed.

The balanced equation for the reaction is given below:

Cr + H₂SO₄ —> CrSO₄ + H₂

From the balanced equation above,

1 mole of Cr reacted to produce 1 mole of CrSO₄.

Finally, we shall determine the number of mole of Cr needed to produce 14.81 moles of CrSO₄. This can be obtained as follow:

From the balanced equation above,

1 mole of Cr reacted to produce 1 mole of CrSO₄.

Therefore, 14.81 moles of Cr will also react to produce 14.81 moles of CrSO₄.

Thus, 14.81 moles of Cr is needed for the reaction.

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3 years ago

The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate

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Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

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3 years ago

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Also, buoyancy force $(F_{b})$ = (volume submerged in water × density of water) + (volume in oil × density of oil)

$(F_{b})$ = $(0.592 V \times \rho) + (1 - 0.592)V \times 1000 g$

= $(0.592 V \times \rho + 408 V)$ g

As, W = V × density of graphite × g

It is given that density of graphite is $2.16 g/cm^{3}$ or 2160 $kg/m^{3}$ .

So, W = 2160 V g

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It is given that mass of flask is 124.8 g.

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For the given chemical equation:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl2(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{Cl_2}\times p_{PCl_3}}{p_{PCl_5}}$

We are given:

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Putting values in above equation, we get:

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Hence, the equilibrium partial pressure of chlorine gas is 0.360 atm

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